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JEE Main & Advanced Mathematics Circle and System of Circles Question Bank Tangent and normal to a circle

  • question_answer
    \[x=7\] touches the circle \[{{x}^{2}}+{{y}^{2}}-4x-6y-12=0\], then the coordinates of the point of contact are           [MP PET 1996]

    A)            (7, 3)                                         

    B)            (7, 4)

    C)            (7, 8)                                         

    D)            (7, 2)

    Correct Answer: A

    Solution :

     Putting\[x=7\], we get \[{{y}^{2}}-6y+9=0\] \[\Rightarrow y=3,\ 3\]                    Hence the point of contact is (7, 3).


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